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2.7. Examples
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Today, we're going to look at how to solve a second-order homogeneous linear differential equation. Let's start with our first example: d²y/dx² - 5 dy/dx + 6y = 0. Can anyone tell me the first step?
We need to find the auxiliary equation, right?
Exactly! The auxiliary equation is m² - 5m + 6 = 0. What type of roots do we expect to find here?
Distinct real roots because the discriminant is positive.
Exactly! So when we solve the auxiliary equation, what roots do we get?
m = 2 and m = 3.
Correct! Therefore, the general solution will be y(x) = C₁e²ˣ + C₂e³ˣ. Great job!
Can we say anything about the behavior of this solution?
Absolutely! Since these are exponential functions, the solutions will grow without bound. This kind of behavior is common in systems where there’s rapid escalation, like in some structures during load applications.
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Now let's tackle our second example: d²y/dx² - 4dy/dx + 4y = 0. Who can provide the auxiliary equation?
It's m² - 4m + 4 = 0.
Correct! And what do we find when we solve it?
There's only one repeated root, m = 2.
That's right. So how does this affect our general solution?
Because it’s a repeated root, the solution will be y(x) = (C₁ + C₂x)e²ˣ.
Perfect! This means that the response of the system will not only grow, but at a linear rate due to the x term in the solution.
Interesting! Does this apply to real-world systems?
Absolutely! Structures might behave like this under certain load conditions, indicating potential failure if not properly managed.
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Now, let’s dive into our third example: d²y/dx² + 4y = 0. Who can write the auxiliary equation?
It's m² + 4 = 0, which gives us complex roots.
Correct! Can anyone state the roots we find?
m = ±2i.
Awesome! What does this signify about the solution?
The general solution will be y(x) = C₁cos(2x) + C₂sin(2x), which indicates oscillatory motion.
Exactly! This is particularly useful when modeling systems experiencing damped vibrations, like in mechanical structures.
So, they will behave like waves?
Correct! An oscillatory response is vital in many engineering applications, especially for systems that need to absorb shock or loads efficiently.
Overview
Short Summary
This section provides examples illustrating the solution of second-order homogeneous linear differential equations with distinct, repeated, and complex roots.
Medium Summary
In this section, we explore three examples of solving second-order linear homogeneous differential equations. Each example highlights different types of roots—distinct, repeated, and complex—demonstrating how to derive general solutions based on the roots' nature.
Detailed Summary
The section begins by solving an equation with distinct real roots, demonstrating how to derive the auxiliary equation and general solution. For repeated roots, the solution method is adjusted, leading to a specific form for the general solution. Lastly, the section addresses equations with complex roots, emphasizing their applications in modeling oscillatory motions. Understanding these examples is crucial for engineering applications where these equations frequently arise.
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Audio Book
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Create a free accountExample 1: Distinct Real Roots
Solve:
d²y/dx² - 5 dy/dx + 6 y = 0
Solution: Auxiliary equation: m² - 5m + 6 = 0 Roots: m = 2, 3 General solution: y(x) = C₁ e²ˣ + C₂ e³ˣ
Detailed Explanation
In this example, we have a second-order linear homogeneous differential equation. We first form the auxiliary equation by replacing y with a function in the form of e^{mx}. This leads us to the characteristic polynomial, which is m² - 5m + 6 = 0. We solve this polynomial to find its roots, which are m = 2 and m = 3. Since we have two distinct real roots, the general solution takes the form y(x) = C₁ e²ˣ + C₂ e³ˣ, where C₁ and C₂ are arbitrary constants determined by initial or boundary conditions.
Examples & Analogies
Imagine a concert with two musicians performing at different pitches. Each musician's sound represents a solution to the differential equation. Just as their unique pitches blend to create harmony, the distinct roots (m = 2 and m = 3) represent two independent solutions that together describe how the system behaves over time.
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Create a free accountExample 2: Repeated Roots
Solve:
d²y/dx² - 4 dy/dx + 4 y = 0
Solution: Auxiliary equation: m² - 4m + 4 = 0 ⇒ Roots: m = 2 (repeated) Solution: y(x) = (C₁ + C₂ x) e²ˣ
Detailed Explanation
In this case, the coefficients of the differential equation lead us to an auxiliary equation with repeated roots. After solving m² - 4m + 4 = 0, we find that m = 2 is repeated. When we have repeated roots in a differential equation, the general solution can be expressed in the form y(x) = (C₁ + C₂ x) e²ˣ. The term C₂ x appears to account for the multiplicity of the root, indicating that there are infinitely many solutions corresponding to the same value of m.
Examples & Analogies
Think of a person running in place but gradually increasing their pace to a steady rhythm. The repeated root symbolizes their constant speed (m = 2), while adding a factor of x reflects their increasing energy and effort over time, creating a unique mix of steady and constant behavior.
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Create a free accountExample 3: Complex Roots
Solve:
d²y/dx² + 4 y = 0
Solution: Auxiliary equation: m² + 4 = 0 ⇒ m = ±2i General solution: y(x) = C₁ cos(2x) + C₂ sin(2x)
Detailed Explanation
Here, the equation leads us to complex roots. By setting up the auxiliary equation m² + 4 = 0, we find that the roots are of the form ±2i. When dealing with complex roots, the general solution involves sine and cosine functions: y(x) = C₁ cos(2x) + C₂ sin(2x). This form reflects oscillatory behavior and is essential in situations such as modeling vibrations in structures.
Examples & Analogies
Consider a swing going back and forth. The swing's motion represents the oscillations captured in our solution. Just like the repetitive motion of the swing traced out in two directions (back and forward), complex roots express how systems behave in oscillatory patterns, critical in understanding things like vibrations in engineering.
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Key concepts
Core takeaways and short definitions to help you quickly recall the key ideas from this section.
- Auxiliary Equation:
A key polynomial derived from a second-order differential equation used to determine the nature of the solutions.
- Distinct Roots:
Solutions to the auxiliary equation that result in exponential functions.
- Repeated Roots:
Solutions that incorporate a polynomial term due to the multiplicity of a root.
- Complex Roots:
Solutions characterized by oscillatory functions, applicable in dynamics and vibrations.
Examples
Step-by-step examples to apply the section's ideas and test your understanding.
Example 1: Distinct Real Roots - Solving d²y/dx² - 5 dy/dx + 6y = 0 yields y(x) = C₁e²ˣ + C₂e³ˣ.
Example 2: Repeated Roots - Solving d²y/dx² - 4dy/dx + 4y = 0 yields y(x) = (C₁ + C₂x)e²ˣ.
Example 3: Complex Roots - Solving d²y/dx² + 4y = 0 yields y(x) = C₁cos(2x) + C₂sin(2x).
Memory aids
Imagine a bridge swaying gently in the breeze; if the roots are complex, it dances with ease, oscillating to and fro, in a rhythm that flows.
For distinct roots, remember 'Advent,’ for different ends, solutions ascend. Repeated roots, think 'Pair to Share,' one solution, but be aware.
Flash Cards
Glossary
Homogeneous Linear Differential Equation
A differential equation where the function and its derivatives are proportional to one another, set to zero.
Auxiliary Equation
A polynomial equation derived from substituting a trial solution into a differential equation, used to find the roots.
Distinct Roots
Roots of the auxiliary equation that are different from one another, leading to exponential solutions.
Repeated Roots
Roots of the auxiliary equation that are the same, leading to a solution that involves a polynomial term.
Complex Roots
Roots that involve imaginary numbers, leading to oscillatory solutions involving sine and cosine functions.